Problem: Simplify the following expression and state the condition under which the simplification is valid: $x = \dfrac{k^2 - 6k + 9}{k^2 - 12k + 27}$
First factor the expressions in the numerator and denominator. $ \dfrac{k^2 - 6k + 9}{k^2 - 12k + 27} = \dfrac{(k - 3)(k - 3)}{(k - 9)(k - 3)} $ Notice that the term $(k - 3)$ appears in both the numerator and denominator. Dividing both the numerator and denominator by $(k - 3)$ gives: $x = \dfrac{k - 3}{k - 9}$ Since we divided by $(k - 3)$, $k \neq 3$. $x = \dfrac{k - 3}{k - 9}; \space k \neq 3$